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Probability, MET2 2024 VCAA 7 MC

A fair six-sided die is repeatedly rolled. What is the minimum number of rolls required so that the probability of rolling a six at least once is greater than 0.95 ?

  1. 15
  2. 16
  3. 17
  4. 18
Show Answers Only

\(C\)

Show Worked Solution

\(X\sim \text{Bi} \left(n, \dfrac{1}{6}\right)\)

\(\text{Pr}(X\geq 1)\) \(>0.95\)
\(1-\text{Pr}(X=0)\) \(>0.95\)
\(-\text{Pr}(X=0)\) \(\geq -0.05\)
\(\text{Pr}(X=0)\) \(< 0.05\quad \left[\text{CAS}:\text{invBinomN}\left(0.05,\dfrac{1}{6},0\right)\right]\)
  \(=17\)

  \(\Rightarrow C\)

Statistics, MET2 2020 VCAA 3

A transport company has detailed records of all its deliveries. The number of minutes a delivery is made before or after its schedule delivery time can be modelled as a normally distributed random variable, `T`, with a mean of zero and a standard deviation of four minutes. A graph of the probability distribution of `T` is shown below.
 

  1. If  `"Pr"(T <= a)=0.6`, find `a` to the nearest minute.   (1 mark)

  2. Find the probability, correct to three decimal places, of a delivery being no later than three minutes after its scheduled delivery time, given that it arrives after its scheduled delivery time.   (2 marks)

  3. Using the model described above, the transport company can make 46.48% of its deliveries over the interval  `-3 <= t <= 2`.
  4. It has an improved delivery model with a mean of `k` and a standard deviation of four minutes.
  5. Find the values of `k`, correct to one decimal place, so that 46.48% of the transport company's deliveries can be made over the interval  `-4.5 <= t <= 0.5`   (3 marks)

A rival transport company claims that there is a 0.85 probability that each delivery it makes will arrive on time or earlier.

Assume that whether each delivery is on time or earlier is independent of other deliveries.

  1. Assuming that the rival company's claim is true, find the probability that on a day in which the rival company makes eight deliveries, fewer than half of them arrive on time or earlier. Give your answer correct to three decimal places.   (2 marks)

  2. Assuming that the rival company's claim is true, consider a day in which it makes `n` deliveries.
    1. Express, in terms of `n`, the probability that one or more deliveries will not arrive on time or earlier.   (1 mark)

    2. Hence, or otherwise, find the minimum value of `n` such that there is at least a 0.95 probability that one or more deliveries will not arrive on time or earlier.   (1 mark)
  3. An analyst from a government department believes the rival transport company's claim is only true for deliveries made before 4 pm. For deliveries made after 4 pm, the analyst believes the probability of a delivery arriving on time or earlier is `x`, where  `0.3 <=x <= 0.7`
  4. After observing a large number of the rival transport company's deliveries, the analyst believes that the overall probability that a delivery arrives on time or earlier is actually 0.75
  5. Let the probability that a delivery is made after 4 pm be `y`.
  6. Assuming that the analyst's belief are true, find the minimum and maximum values of `y`.   (2 marks)

Show Answers Only

  1. `a= 1\ text(minute)`
  2. `0.547`
  3. `k=-1.5, -2.5`
  4. `0.003`
  5.  i. `1-0.85^n`
  6. ii. `19`
  7. `2/3`

Show Worked Solution

a.   `T\ ~\ N(0, 4^2)`

`text(Solve (by CAS): Pr)(T<=a) = 0.6`

`:. a= 1\ text(minute)`
 

b.    `text{Pr}(T <= 3∣T > 0)` `=(text{Pr}(0 < T <= 3))/(text{Pr}(T > 0))`
    `=(0.27337 dots)/(0.5)`
    `=0.547\ \ text{(to 3 d.p.)}`

 

c.   `text(Given)\ \ text{Pr}(-3 <= T <= 2) = 0.4648`

`sigma = 4 text{minutes}`

`=> \ text{Pr}(-4.5 <= T – 1.5 <= 0.5) = 0.4648`

`=> k=-1.5`

`text(By symmetry of the normal distribution)`

`text{Pr}(-2 <= T <= 3) = text{Pr}(-3 <= T <= 2) = 0.4648`

`=> \ text{Pr}(-4.5 <= T – 2.5 <= 0.5) = 0.4648`

`=> k=-2.5`

`:. k=-1.5, -2.5`

 

d.   `text{Let}\ \ X\ ~\ text{Bi}(8, 0.85)`

`text(Solve (by CAS):)`

`text{Pr}(X<=3) = 0.003\ \ text{(to 3 d.p.)}`
 

e.i.   `text{Pr(at least 1 delivery is late)}`

`= 1-\ text{Pr(all deliveries are on time)}`

`=1-0.85^n`
 

e.ii.   `text{Solve for}\ n:`

`1-0.85^n` `<0.95`  
`n` `>18.43…`  

 
`:.n_min=19`
 

f.   `text{Pr(delivery made after 4pm)} = y`

`=>\ text{Pr(delivery made before 4pm)} = 1-y`

`0.85(1-y)+xy` `=0.75`  
`y` `=-(0.1)/(x-0.85)`  
  `=(2)/(17-20 x)`  

 
`text(Given ) 0.3<=x<=0.7:`

`y_min = (2)/(17-20 xx 0.3) = 2/11`

`y_max = (2)/(17-20 xx 0.7) = 2/3`

Probability, MET2 2008 VCAA 14 MC

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is

A.     `8`

B.     `9`

C.   `10`

D.   `11`

E.   `12`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ X = text(Number of heads,)`

`X ∼ text(Bi) (n, 1/2)`

`text(Pr) (X = n)` `< 0.0005`
`((n), (n)) (1/2)^n (1/2)^0` `< 0.0005`
`n` `> 10.97`

`:. n_min = 11`

`=>   D`

Probability, MET2 2010 VCAA 2

Shoddy Ltd produces statues that are classified as Superior or Regular and are entirely made by machines, on a construction line. The quality of any one of Shoddy’s statues is independent of the quality of any of the others on its construction line. The probability that any one of Shoddy’s statues is Regular is 0.8.

Shoddy Ltd wants to ensure that the probability that it produces at least two Superior statues in a day’s production run is at least 0.9.

Calculate the minimum number of statues that Shoddy would need to produce in a day to achieve this aim.  (3 marks)

Show Answers Only

`18`

Show Worked Solution

`text(Solution 1)`

`text(Let)\ \ X = text(Number of superior statues),`

♦♦ Mean mark 27%.

`X∼\ text(Bi) (n, 0.2)`

`n=18`

 

`text(Solution 2)`

`text(Pr) (X >= 2)` `>= 0.9`
`1 – text(Pr) (X = 0) – text(Pr) (X = 1)` `>= 0.9`
`1 – 0.8^n – ((n),(1))(0.2)^1 (0.8)^(n – 1)` `>= 0.9`
`n` `>= 17.9`
`:. n_min` `= 18`

Probability, MET2 2014 VCAA 4*

Patricia is a gardener and she owns a garden nursery. She grows and sells basil plants and coriander plants.

The heights, in centimetres, of the basil plants that Patricia is selling are distributed normally with a mean of 14 cm and a standard deviation of 4 cm. There are 2000 basil plants in the nursery.

  1. Patricia classifies the tallest 10 per cent of her basil plants as super.
  2. What is the minimum height of a super basil plant, correct to the nearest millimetre?   (1 mark)

Patricia decides that some of her basil plants are not growing quickly enough, so she plans to move them to a special greenhouse. She will move the basil plants that are less than 9 cm in height.

  1. How many basil plants will Patricia move to the greenhouse, correct to the nearest whole number?   (2 marks)

The heights of the coriander plants, `x` centimetres, follow the probability density function  `h(x)`,

`h(x) = {(pi/100 sin ((pi x)/50), 0 < x < 50), (\ \ \ \ \ \0, text(otherwise)):}`

  1. State the mean height of the coriander plants.   (1 mark)

Patricia thinks that the smallest 15 per cent of her coriander plants should be given a new type of plant food

  1. Find the maximum height, correct to the nearest millimetre, of a coriander plant if it is to be given the new type of plant food.   (2 marks)

Patricia also grows and sells tomato plants that she classifies as either tall or regular. She finds that 20 per cent of her tomato plants are tall.

A customer, Jack, selects `n` tomato plants at random.

  1. Let `q` be the probability that at least one of Jack’s `n` tomato plants is tall.
  2. Find the minimum value of `n` so that `q` is greater than 0.95.   (2 marks)

Show Answers Only

  1. `191\ text(mm)`
  2. `211\ text(plants)`
  3. `25\ text(cm)`
  4. `127\ text(mm)`
  5. `14\ text(plants)`

Show Worked Solution

a.   `text(Let)\ \ X = text(plant height,)`

♦ Mean mark 43%.

`X ∼\ text(N)(14,4^2)`

`text(Pr)(X > a)` `= 0.1`
`a` `= 19.1\ text(cm)quadtext([CAS: invNorm)\ (.9,14,4)]`
  `=191\ text{mm  (nearest mm)}`

 

`:.\ text(Min super plant height is 191 mm.)`

 

b.   `text(Pr)(X < 9) = 0.10565…\ qquadtext([CAS: normCdf)\ (−∞,9,14,4)]`

`:.\ text(Number moved to greenhouse)`

`= 0.10565… xx 2000`

`= 211\ text(plants)`

 

c.    `text(E)(X)` `= int_0^50  (x xx pi/100 sin((pix)/50))dx`
    `= 25\ text(cm)`

 

d.    `text(Solve:)\ \ int_0^a h(x)\ dx` `= 0.15\ \ text(for)\ \ a ∈ (0,50)`

♦♦ Mean mark 33%.

`:.a` `=12.659…\ text(cm)`
  `=127\ text{mm  (nearest mm)}`

 

e.   `text(Let)\ \ Y =\ text(Number of tall plants,)`

♦♦ Mean mark 30%.

`Y ∼\ text(Bi) (n,0.2)`

`text(Pr)(Y >= 1)` `> 0.95`
`1-text(Pr)(Y = 0)` `> 0.95`
`0.05` `> 0.8^n`
`n` `> 13.4\ \ text([by CAS])`

 

`:. n_text(min) = 14\ text(plants)`